$\begin{array}{1 1} 12.2\;m/s \\ 3.5\;m/s \\ 4.8\;m/s\\ 13.1\;m/s\end{array} $

Answer :13.1 m/s

In this case, the force that enables the car to remain in its circular path is the force of static friction.

$f_s =m \large\frac{v^2}{r}$

The maximum speed the car can have around the curve is the speed at which it is on the verge of skidding outward.

$v_{max} =\sqrt { \large\frac{f_{s, max}r}{m}}$

$\qquad= \sqrt {\large\frac{\mu_s m gr}{m}}$

$\qquad =\sqrt {\mu_s gr}$

$\qquad=\sqrt { (0.500)(9.80\;m/s^2)(35.0)}$

$\qquad= 13.1 \;m/s$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...